E1.2 - Three sets and subset notation (Extended) | Cambridge IGCSE International Mathematics

A person can belong to three clubs at once, and a whole set can sit inside another set. Extend the set language of E1.2 to distinguish elements from sets, use subset notation and organise three overlapping groups. You will also interpret a set defined by a rule for coordinate pairs.

Membership and empty set

If A is the set of even natural numbers below 8, is 0 in A? Natural numbers include zero in this course, and zero is even because $0=2\times0$ .

Membership

The notation $x\in A$ means “ $x$ is an element of A”. The notation $x\notin A$ means “ $x$ is not an element of A”. Use membership to compare an individual object with a set.

For $A=\{0,2,4,6\}$ , zero is an element, so $0\in A$ . The braces describe the collection; the numbers inside are its individual elements.

Worked example 1: membership statements

Question. Let $A=\{0,2,4,6\}$ . Complete $2\ \ldots\ A$ and $3\ \ldots\ A$ using $\in$ or $\notin$ , and find $n(A)$ .

Plan. Check whether each object occurs in the list, then count the distinct elements.

\begin{aligned} 2&\in A &&\text{2 is listed}\\ 3&\notin A &&\text{3 is not listed}\\ n(A)&=4 &&\text{0, 2, 4 and 6} \end{aligned}

Answer. $2\in A$ , $3\notin A$ , and $n(A)=4$ .

Check. Every listed number is even and below 8; 3 fails the even-number condition.

A condition can select no objects at all.

Empty set

The empty set, written $\varnothing$ or $\{\}$ , has no elements. Thus $n(\varnothing)=0$ . The set $\{0\}$ is different: it has the single element 0.

The next example changes a nonempty list to a condition that no allowed number satisfies.

Worked example 2: a set with no elements

Question. Let $B=\{x\mid x\text{ is a natural number and }4<x<5\}$ . Write B using set notation, find $n(B)$ , and state whether $0\in B$ .

Plan. Check for natural numbers strictly between 4 and 5.

There is no whole number in that interval, so

\begin{aligned} B&=\varnothing\\ n(B)&=0\\ 0&\notin B \end{aligned}

Answer. $B=\varnothing$ , $n(B)=0$ , and $0\notin B$ .

Check. The consecutive natural numbers 4 and 5 are both excluded by the strict bounds. Zero also fails $4<x$ .

An empty overlap is written $A\cap B=\varnothing$ , while its count is $n(A\cap B)=0$ . Keep the set and the count distinct.

Watch out

The statement $\varnothing=0$ confuses a set with a number. Write $n(\varnothing)=0$ . Also, $\{0\}\ne\varnothing$ , because $n(\{0\})=1$ .

Use the definitions to choose the correct pair of statements.

Subsets

Membership asks about one object. A subset statement asks whether every element of one set also belongs to another set.

Subset and not a subset

The notation $A\subseteq B$ means A is a subset of B: every element of A is also an element of B. Equality is allowed. The notation $A\nsubseteq B$ means at least one element of A is not in B.

To establish a subset relationship for finite sets, check all elements. To disprove it, one counterexample is enough. For example, $\{1,4\}\nsubseteq\{1,2,3\}$ because 4 is missing from the second set.

Worked example 3: a subset defined by conditions

Question. Let $U=\{0,1,2,\ldots,12\}$ , $A=\{x\mid x\text{ is a multiple of }4\}$ , and $B=\{x\mid x\text{ is even}\}$ . Complete $A\ \ldots\ B$ using $\subseteq$ or $\nsubseteq$ . Give a reason.

Plan. List the allowed elements and check every element of A against B.

\begin{aligned} A&=\{0,4,8,12\}\\ B&=\{0,2,4,6,8,10,12\} \end{aligned}

All four elements of A occur in B.

Answer. $A\subseteq B$ , because every allowed multiple of 4 is even.

Check. Each element of A is twice a whole number: $0=2\times0$ , $4=2\times2$ , $8=2\times4$ , $12=2\times6$ .

Now add a bound to B's rule. A general fact about multiples is not enough if the allowed ranges differ.

Worked example 4: a counterexample to a subset claim

Question. Keep $U=\{0,1,2,\ldots,12\}$ and $A=\{0,4,8,12\}$ , but let $C=\{x\mid x\text{ is even and }x<10\}$ . Is $A\subseteq C$ ? Give a reason.

Plan. Look for an element of A that fails C's added bound.

\begin{aligned} C&=\{0,2,4,6,8\}\\ 12&\in A\quad\text{but}\quad12\notin C \end{aligned}

Answer. $A\nsubseteq C$ , because 12 is in A but not in C.

Check. The other elements 0, 4 and 8 are in C, but “every” fails as soon as 12 fails.

Every set is a subset of itself, so $A\subseteq A$ . The empty set is a subset of every set: it has no element that could fail the condition. Neither statement says that an individual number is a set.

Watch out

For $B=\{2,5,8\}$ , write $5\in B$ for an element and $\{5\}\subseteq B$ for a one-element set. The two left-hand objects have different roles.

Choose a counterexample if the subset statement is false.

Three-set diagrams

Three overlapping sets divide the universe into eight disjoint regions: three “only” regions, three “exactly two” regions, one triple overlap and one outside all three. Put each object or count in exactly one of these regions.

Three-set intersection

The set $A\cap B\cap C$ contains elements in all three sets. The set $A\cap B$ also includes this triple overlap. To select A and B but not C, use $A\cap B\cap C'$ .

For numerical survey data, fill the triple overlap first. Then subtract it from each pairwise total before filling the single-set regions.

Worked example 5: complete a three-set survey

Question. In a group of 40 people, $n(A)=21$ , $n(B)=19$ , $n(C)=16$ , $n(A\cap B)=9$ , $n(A\cap C)=7$ , $n(B\cap C)=6$ , and $n(A\cap B\cap C)=4$ . Find all eight region counts.

Plan. Work from the centre outwards so no person is counted twice.

\begin{aligned} \text{all three}&=4\\ \text{A and B only}&=9-4=5\\ \text{A and C only}&=7-4=3\\ \text{B and C only}&=6-4=2\\ \text{A only}&=21-5-3-4=9\\ \text{B only}&=19-5-2-4=8\\ \text{C only}&=16-3-2-4=7\\ \text{none}&=40-(9+8+7+5+3+2+4)=2 \end{aligned}

Answer. A only 9, B only 8, C only 7; A and B only 5, A and C only 3, B and C only 2; all three 4; none 2.

Check. Each circle has its stated total: $9+5+3+4=21$ , $8+5+2+4=19$ , $7+3+2+4=16$ . All eight regions sum to 40.

The completed diagram distinguishes a pairwise intersection from its “only” region.

[DIAGRAM: asset_name: Eight regions in a three-set survey; asset_slug: three_sets_counts; caption: The centre is included in each pairwise intersection. For example, n of A intersection B is 5 plus 4.; description: Three overlapping circles A, B and C lie in a universal rectangle. The A-only, B-only and C-only counts are nine, eight and seven. The A-and-B-only, A-and-C-only and B-and-C-only counts are five, three and two. The triple overlap contains four and the outside region contains two.]
The centre is included in each pairwise intersection. For example, n of A intersection B is 5 plus 4.

A set can contain coordinate pairs. The pair $(x,y)$ has first coordinate $x$ and second coordinate $y$ ; its order matters. The rule $y=x+1$ selects pairs whose second number is one greater than the first. For example, $(2,3)$ satisfies it because $3=2+1$ .

Worked example 6: sets defined by coordinate rules

Question. Let $U=\{(0,0),(0,1),(1,0),(1,2),(2,2),(2,3)\}$ . Define $A=\{(x,y)\mid x=0\}$ , $B=\{(x,y)\mid y=x+1\}$ , and $C=\{(x,y)\mid y\ge2\}$ , using only pairs in $U$ . Place the pairs into the Venn regions and find $A\cap B$ and $B\cap C$ .

Plan. Test each pair against all three conditions; use the complete pair as one element.

Pair	In A?	In B?	In C?	Region
$(0,0)$	Yes	No	No	A only
$(0,1)$	Yes	Yes	No	A and B only
$(1,0)$	No	No	No	Outside all three
$(1,2)$	No	Yes	Yes	B and C only
$(2,2)$	No	No	Yes	C only
$(2,3)$	No	Yes	Yes	B and C only

Answer. The placements are shown in the table. $A\cap B=\{(0,1)\}$ and $B\cap C=\{(1,2),(2,3)\}$ . The B-only, A-and-C-only and triple regions are empty.

Check. All six original pairs appear once. For the two elements of $B\cap C$ , the second coordinate is both at least 2 and one greater than the first.

The mathematical rule changes in this example, but the Venn decision stays the same: test each condition and place the element once.

[DIAGRAM: asset_name: Coordinate pairs in three sets; asset_slug: three_sets_pairs; caption: Treat each coordinate pair as one element and test its two coordinates against the set rules.; description: A three-set Venn diagram for the six coordinate pairs in the preceding table. A-only has zero comma zero, A-and-B-only has zero comma one, B-and-C-only has one comma two and two comma three, C-only has two comma two, and outside all three has one comma zero. B-only, A-and-C-only and the centre are labelled empty.]
Treat each coordinate pair as one element and test its two coordinates against the set rules.

Watch out

“In A and B” includes anyone also in C. “In A and B only” excludes C. Subtract the centre from pairwise totals before placing the exclusive pair counts.

Use a centre-out calculation for this smaller survey.

Quick Recap

Use $\in$ and $\notin$ for elements; $\varnothing$ is a set with no elements.
Use $\subseteq$ for a subset, allowing equality; one missing element proves $\nsubseteq$ .
Fill three-set counts from the centre outwards, then check the complete universe.
A coordinate pair is one ordered element; test its coordinates against each set's defining rule.

Three clubs can overlap without confusion when each person is placed in just one of the eight regions.

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E1.2 - Three sets and subset notation (Extended)

Membership and empty set

Subsets

Three-set diagrams

Quick Recap

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